## Euclidean Space from a Sphere.

### May 10, 2010

Just a quick note here; I wanted to mention a kind of “trick” when doing some algebraic topology problems.  A few problems from Hatcher reminded me of this, and it’s kind of strange that I didn’t bring it up before.

cut cut cut.

Let’s first (informally) define stereographic projection.  I’m just going to define it for the circle, since it’s similar in higher dimensions.

Take a circle and consider the “north pole” or the highest point on the circle.  What we’re going to do is “project” the points on the circle to points on the real line.  So, consider a line from the north pole that cuts through some other point in the circle, say P.  Then there exists a point P’ on the real line such that this line intersects the real line at P’.  This picture really shows what I’m talking about (stolen from wikipedia).  Notice, when you look at this picture, that if you draw another line from the north pole and intersect it at some other point K on the circle, then K’ on the real line will be different from P’.  Try drawing this yourself to see.

Now, what’s sweet about this is: for every point on the circle, except the north pole, there exists a point on the real line that corresponds with that point on the circle via a line from the north pole to that point on the circle.  So, for example, P’ in the picture corresponds to P, and vice versa.  That means that if we made a map from the points on the circle to their corresponding points on the real line, it would be bijective (notice that we do not consider the north pole to the north pole because this would be a horizontal line which is parallel to the x-axis).  Similarly, it’s not difficult to show (but I won’t because I’m pretty lazy) that this is a continuous map and has a continuous inverse (if you write down the formula for computing what P’ is explicitly, this becomes obvious).  This means that there is a homeomorphism from the circle minus the north pole to the real line; this means that the circle minus a point is homeomorphic to the real line!

This proof can be extended for any n-sphere, and we obtain this extremely useful theorem:

Thm:  For any n, $S^{n}-\{N\}$ is homeomorphic to ${\mathbb R}^n$ (where $S^n$ is the n-sphere and $N$ is the point at the “north pole”).

Let’s do one example of how to apply this to a problem:

Problem: Show that $S^n$ is not homeomorphic to $S^m$ if $m\neq n$.  (Assume that we know all the theorems from previous posts.)

This problem can be done pretty easily if we remember that ${\mathbb R}^n$ is not homeomorphic to ${\mathbb R}^m$ if $m\neq n$.  That simplifies a lot.

So suppose that $S^n$ was homeomorphic to $S^m$ where $m\neq n$.  Then they would still be homeomorphic if we removed a point; say, the north pole.  Removing the north pole, we note that $S^n$ without the north pole is homeomorphic to ${\mathbb R}^n$ and $S^m$ without the north pole is homeomorphic to ${\mathbb R}^m$.  This implies that ${\mathbb R}^n$ is homeomorphic to ${\mathbb R}^m$, but this contradicts our theorem in the last post, since we’ve assumed \$m\neq n\$.  Contradiction.  $\Rightarrow\Leftarrow$.  It follows that $S^n$ is not homeomorphic to $S^m$ where $m\neq n$$\Box$